4 playing cards are required to be picked up .what is the probability that card is picked up from each suit ??
Please help.
probability
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 126
- Joined: Sun Jun 24, 2012 10:11 am
- Location: Chicago, IL
- Thanked: 36 times
- Followed by:7 members
Picking the cards in sequence are conditional events so multiply their probabilities at the end. Manhattan GMAT guides call that 'the domino' effect.
Picking the first card, can be any card, 52 out of 52: probability = 52/52=1.
Picking the second card, can be anything except the suit of card 1 -> picking amongst 3 suits = 3 x 13 = 39 cards out of 51: probability = 39/51.
Picking the third card, can be anything except the suits of card 1 and card 2 -> picking amongst 2 x 13 = 26 cards out of 50: probability = 26/50.
Picking the forth card, can be anything except the suits of cards 1,2,3 -> picking amongst 13 cards out of 49: probability = 13/49.
Multiply the 4 probabilities to get total probability P = 1 x 39/51 x 26/50 x 13/49 and simplify the fraction.
Picking the first card, can be any card, 52 out of 52: probability = 52/52=1.
Picking the second card, can be anything except the suit of card 1 -> picking amongst 3 suits = 3 x 13 = 39 cards out of 51: probability = 39/51.
Picking the third card, can be anything except the suits of card 1 and card 2 -> picking amongst 2 x 13 = 26 cards out of 50: probability = 26/50.
Picking the forth card, can be anything except the suits of cards 1,2,3 -> picking amongst 13 cards out of 49: probability = 13/49.
Multiply the 4 probabilities to get total probability P = 1 x 39/51 x 26/50 x 13/49 and simplify the fraction.
Skype / Chicago quant tutor in GMAT / GRE
https://gmat.tutorchicago.org/
https://gmat.tutorchicago.org/